PROBLEM: Alice, Bob, Catherine, Doug, and Edna are randomly assigned seats at a circular table in a perfectly circular room. Assume that rotations of the table do not matter, so there are exactly 24 possible outcomes in the sample space.
Bob and Catherine are married. Doug and Edna are married. When people are married they love to sit beside each other.
Let T denote the event that Bob and Catherine are sitting next to each other. Let U be the event that Alice and Bob are sitting next to each other. Are events T and U independent? Why? Justify your answer.
My QUESTIONS:
1. Problem statement says couples love to sit beside each other. Why did author include this statement? Does he mean by "love to sit beside each other" that "they are always sitting beside each other" or he is just trying to confuse you ?
2. How to start solving the problem. I mean I can't even think of with what I should starts ;(
If by "love to" author meant they are "always sitting together" then nothing matters and $P(T) = 1$ and Alice has choice to sit next to rest 4 people: $P(U) = \frac14$
Since they always sit together and Alice got its own wish to sit anywhere, no one is forcing her, they are independent events. This is the reasoning I could come up with.
$1)$ The author means nothing relevant when he says that. It's just an an inane piece of information meant to provide some idea of a context for the actual problem statement. It has nothing to do with solving the problem, i.e. even if the author had not put this line, the solution to the problem would not have changed one iota.
$2)$ I am going to give a mathematical proof here. There might be a logical way to do this too. To find whether the events are independent or not, we are going to use the fact that if two events are independent, then
$$P(A\cap B) = P(A)P(B)$$
Now to calculate $P(T)$ and $P(U)$.
Starting with $T$, we see that Bob and Catherine are sitting together. So we combine them in one unit and calculate number of ways of seating them around the table (BC),E,D,A (Here letters represent the initials of the names of people). Since there are 4 entities, there are $3!$ ways to do it. Also, we can shift Bob and Catherine around (as in (BC) and (CB)) in two ways giving us
$$|T| = 3!\cdot2!$$ $$P(T) = \frac{3!\cdot2!}{4!}=\frac12$$
Applying a similar logic for $U$, we see that Bob and Alice are sitting together. So we combine them in one unit and calculate number of ways of seating them around the table (BA),E,D,C (Here letters represent the initials of the names of people). Since there are 4 entities, there are $3!$ ways to do it. Also, we can shift Bob and Alice around (as in (BA) and (AB)) in two ways giving us
$$|U| = 3!\cdot2!$$ $$P(U) = \frac{3!\cdot2!}{4!}=\frac12$$
Now we calculate $P(T\cap U)$. For this we put $T$ and $U$ together, i.e. both Alice and Catherine are sitting together with Bob. So we combine the three of them in one unit and try to seat 3 entities around a table (CBA),D,E. This can be done in $2!$ ways. Also Catherine and Alice can switch places inside the entity giving us a multiplicative factor of $2$. Now the total number of ways is
$$|T\cap U| = 2!\cdot 2!$$ $$P(T\cap U) = \frac{2!\cdot 2!}{4!} = \frac16$$
We can see that $P(T\cap U) \ne P(T)P(U)$, meaning they are not independent.
I am attaching a few links regarding independent events in probability and circular permutations for your understanding
https://brilliant.org/wiki/probability-independent-events/
https://www.toppr.com/guides/business-mathematics-and-statistics/permutations-and-combinations/permutations-and-circular-permutation/